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Quantum degeneracy problem, electron on a ring

  1. Aug 21, 2012 #1
    1. The problem statement, all variables and given/known data

    yvOtC.png

    2. Relevant equations

    Below

    3. The attempt at a solution

    So this is a lot like the infinite square well, except periodic. If S is an arc length, then [itex]S = \theta R[/itex] so [itex]\frac{d^2}{dS^2} = \frac{1}{R^2}\frac{d^2}{d\theta^2}[/itex], which is more convenient to use in the hamiltonian. So for the hamiltonian I get:

    [tex]H = \frac{-\hbar^2}{2m}\frac{1}{R^2}\frac{d^2}{d\theta^2} + V_0[/tex]

    With Schrodinger's equation, I get

    [tex]\frac{d^2 \psi^2}{d\theta^2} = -\frac{2mR^2(E - V_0)}{\hbar^2}\psi[/tex]

    Which gives solutions of the form [itex]\psi = \frac{1}{\sqrt{2\pi}}e^{\pm i k \theta}[/itex], where [itex]k = \sqrt{\frac{2mR^2(E - V_0)}{\hbar^2}}[/itex].

    Then, because it's a ring, we need [itex]\psi(x + 2\pi) = \psi(x)[/itex] for any x, which gives us the requirement that k is an integer. So our energy levels are [itex]E = \hbar^2 k^2/2mR^2 + V_0[/itex], and it seems like they have a degeneracy of 4 because we have two functions for each k with the same energy, and then for each of them, the electron's spin can be up or down. Is that right?

    As for part (b), I have no idea... Benzene has 6 free electrons, so according to my degeneracy, it completely fills the first energy level, and then there are 2 electrons in the 2nd energy level. Ok...then they ask about a compound with 4 electrons. This seems like it just fills the first energy level, but that seems too simple and stupid to be right.

    Can anyone help me out?

    Thanks!
     
  2. jcsd
  3. Aug 21, 2012 #2

    TSny

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    k is an integer including zero. What is the degeneracy of the energy level corresponding to k = 0?
     
  4. Aug 22, 2012 #3
    Well I guess that just has a degeneracy of 2 due to the spin, right?

    Any idea on the aromatic thing? I'm totally clueless about that...
     
  5. Aug 22, 2012 #4
    Ah, I think I see, after what you just said and reading the wiki article on Huckel's Rule... The number of states for k =/= 0 is 4 for each energy level. For k = 0 it's 2. So like Huckel's Rule, for energy level n, there are 4n + 2 states. So benzene is aromatic because it has 6 electrons, so fully completes the n = 1 energy level. The other compound has 4 however, so it fills the n = 0 level but only half fills the n = 1 level, so it's not stable (though from the little I know of chemistry, I thought that just meant it's more reactive, not less stable).
     
  6. Aug 22, 2012 #5

    TSny

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    I think your analysis is now correct. I'm not very clear on the reactive vs. stable interpretation either. Maybe someone can clarify it.
     
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